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Mugwump101
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Factor Completely
« on: Apr 20th, 2006, 12:05pm » |
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Using the equation, 8x3+27 (as in 8 X cubed plus 27) , one should factor it out.
Is it possible to factor this using the Quadratic Equation? If so, how and for now, just explain what should I do and what steps should I follow?
Thank you for your time...
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| « Last Edit: Apr 20th, 2006, 3:01pm by Icarus » |
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pex
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Re: Factor Completely
« Reply #1 on: Apr 20th, 2006, 12:57pm » |
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This is what I would do: it is easy to find a solution (say, x = a) to 8x3 + 27 = 0. This means that it is possible to factor 8x3 + 27 = (x - a)(px2 + qx + r).For the second factor you can use the Quadratic Formula.
Here is the factorization - I don't know if you would like to try to find it youself first, so I will hide it: a = -3/2, therefore 8x3 + 27 = (x + 3/2)(8x2 - 12x + 18) = (2x + 3)(4x2 - 6x + 9). From the Quadratic Formula, we only need the discriminant: (-6)2 - 4*4*9 < 0, so the second factor cannot be factored further into polynomials with real coefficients.
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Mugwump101
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Re: Factor Completely
« Reply #2 on: Apr 20th, 2006, 1:36pm » |
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Thank you so much!! But heh honestly I didn't understand what do you did.
P, q, and r are variables? If so, what do they represent? ANd how are you supposed to get 27 and 8 from the equation.. (x-a)(px^2 + qx + r).
Can you write it out in steps like 1), 2) 3) and explain?
Thank so much again for your time!!
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pex
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Re: Factor Completely
« Reply #3 on: Apr 20th, 2006, 2:45pm » |
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I'm sorry, it's really hard to guess the level of math knowledge a poster has...
Anyway, I first find a value that makes 8x3 + 27 = 0 - I called it "a" in my previous post. We have
8x3 + 27 = 0
8x3 = -27
(2x)3 = (-3)3
so an x satisfying 2x = -3 will do - that is, we can set a = -3/2.
This means that it must be possible to factor x - (-3/2) = x + 3/2 from 8x3 + 27. The remaining factor must be quadratic in x, say, px2 + qx + r, where p, q and r are to be determined.
To find p, q and r, write
(x + 3/2)(px2 + qx + r) = 8x3 + 27
px3 + qx2 + rx + (3/2)px2 + (3/2)qx + (3/2)r = 8x3 + 27
px3 + ((3/2)p + q)x2 + ((3/2)q + r)x + (3/2)r = 8x3 + 27
So p, q and r should satisfy p = 8, (3/2)p + q = 0, (3/2)q + r = 0 and (3/2)r = 27: this way we will have px3 + ((3/2)p + q)x2 + ((3/2)q + r)x + (3/2)r = 8x3 + 0x2 + 0x + 27 = 8x3 + 27. Solving yields p = 8 (d'oh), r = 18 (from (3/2)r = 27) and q = -12 (by substituting p = 8 into (3/2)p + q = 0).
This implies that 8x3 + 27 = (x + 3/2)(8x2 - 12x + 18). It looks better if we transfer a factor of two from the quadratic factor to the linear factor: (x + 3/2)(8x2 - 12x + 18) = (x + 3/2) * 2 * (4x2 - 6x + 9) = (2x + 3)(4x2 - 6x + 9).
Now we try to factor 4x2 - 6x + 9. By the Quadratic Formula, this is possible only if the determinant (-6)2 - 4 *4 * 9 were positive or zero. However, it equals -108, so 4x2 - 6x + 9 cannot be factored.
Therefore, 8x3 + 27 = (2x + 3)(4x2 - 6x + 9) is as good as it gets. If I still haven't been specific enough (which often seems to be the case - I'm not particularly good at explaining things), please don't hesitate to ask.
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Icarus
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Re: Factor Completely
« Reply #4 on: Apr 20th, 2006, 3:43pm » |
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Somewhat easier to follow, I hope, though it is essentially the same as what pex has said:
First find a root like pex:
8x3 + 27 = 0
x3 = -27/8
x = cube root (-27/8) = -3/2
So a = -3/2 is one root of the polynomial.
Any root a of polynomial corresponds to a factor of (x - a), so (x - -3/2) = (x + 3/2) is one factor of 8x3 + 27. To make things easier (at least to show in this forum), multiply by 2: (2x + 3).
Now do polynomial long division (or better, synthetic division if you've learned about it yet).
4x2 - 6x + 9
________________________
2x+3 | 8x3 + 0x2 + 0x + 27
- 8x3 + 12x2
______________________
-12x2 + 0x + 27
- -12x2 -18x
______________________
18x + 27
- 18x + 27
______________________
0
So 8x3 + 27 = (2x+3)(4x2 - 6x + 9).
Now, depending on where you are in your "mathematical journey", this either does NOT factor further, or else you can use the Quadratic formula to factor 4x2 - 6x + 9 into 2 linear factors.
What does that mean? If you do not yet know what a "complex number" is, then 4x2 - 6x + 9 cannot be factored.
If you do know what one it, then you can use the quadratic formula to find the two roots r,s of 4x2 - 6x + 9, to factor it as 4(x - r)(x - s).
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| « Last Edit: Apr 24th, 2006, 5:35pm by Icarus » |
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pex
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Re: Factor Completely
« Reply #5 on: Apr 20th, 2006, 4:01pm » |
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Just to be complete: allowing complex numbers and writing V- for the square root symbol, it works out to
8x3 + 27 = (1/4)(2x + 3)(4x - 3 - 3i V-3)(4x - 3 + 3i V-3)
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Mugwump101
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Re: Factor Completely
« Reply #6 on: Apr 21st, 2006, 5:21pm » |
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Oh!!! I understand the 8x ^3 + 27 = (2x + 3)(4x^2 - 6x + 9) part. Someone taught me this method today and I finally got it. You see, this problem was a little ahead of my grade, I'm taking Math B(Advanced Algebra I guess other states call it? With proofs and stuff) But this is a Calculas formula. I looked it up and found more information on it. Thank you so much!!
The furthur analysis is so interesting! With imaginary #s and of the like, I want to learn how to do that too but I guess a little later.
Thank you so much again!! I got it now  ! 
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Icarus
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Re: Factor Completely
« Reply #7 on: Apr 21st, 2006, 7:52pm » |
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You're quite welcome. But I do have to say that 8x3 + 27 = (2x+3)(4x2 - 6x + 9) is strictly algebra, not calculus. Perhaps whoever told you this meant that the formula is useful in calculus. However the formula itself is derived using only algebra, and its applications - even in calculus - are still just algebra.
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Mugwump101
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Re: Factor Completely
« Reply #8 on: Apr 23rd, 2006, 7:33pm » |
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I see, sorry about that.
Okay, for the formula is called a synthetic division with
a^3+b^3=(a-b)(a^2-ab+b) ??
I was given this formula. One thing I don't understand the a^3 and b^3.... Could you explain that to me?
Sorry about all this.
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towr
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Re: Factor Completely
« Reply #9 on: Apr 24th, 2006, 1:21am » |
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I think it should be
a^3+b^3=(a+b)(a^2-ab+b^2)
And with a=2x and b=3 you get the formula from your first post, and the answer.
8x^3+27=(2x+3)(4x^2-6x+9)
Basicly it's just a template, giving a general solution. Match your specific problem to it (identifying in this case that a must be 2x, and b must be 3), and you get your specific solution.
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Icarus
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Re: Factor Completely
« Reply #10 on: Apr 24th, 2006, 5:27pm » |
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Mugwump101, it is clear that you have some serious confusion about the concepts of variables and parameters. This is not uncommon for someone who is at your level. But it needs to be dealt with now, or you will find yourself just getting more and more confused. I am going to try to explain some of this, but I doubt I can make it clear to you over this forum. I strongly suggest that you speak to your math teacher about it, or someone who is able to tutor you.
A "parameter" is a letter used to represent a number (or other expression) that we consider "fixed". That is, in an application, we expect to see a number there instead of a letter (or a different expression).
A "variable" is a letter used as a placeholder where we will consider the effects of putting many numbers. We use variables to express relationships.
By convention, we commonly use letters from the end of the alphabet for variables, and letters from the beginning of the alphabet for parameters. But this is a convention that is often broken, so it is mostly context that tells us which is which. In general though, x, y, z, and t are almost always variables, while a, b, c, d are almost always parameters.
What does that mean? Consider the expression (x + a). In this, x is a variable, while a is parameter. When dealing with an "instance" of this expression, I would expect to still see the x, but to see a number or something else in place of a. For instance; "x + 2", "x+5/2", "x - 6" (= x + (-6) ), "x + sqrt(2)" are all examples of expressions of the form "x + a". Instead of a number for a, we might also have another expression "x + (2y-3)" is another expression of the form "x + a".
What about the x, then? The variable also gets replaced by values, but in this case, we are not interested in considering x to be replaced by some specific value, but rather as serving as a placeholder for all values, which may be put in one at a time to see the results.
You may not think that this is a valid distinction, and if you don't, you are correct. The only difference between parameters and variables is how and when we plan to replace them with values. But valid or not, it is a very useful distinction.
Consider the formula that you gave for factoring the sum of two cubes (with the correction that towr points out):
a3 + b3 = (a + b)(a2 - ab + b2).
In this formula, a and b are parameters. If you substitute any valid expression for a in every place it occurs, and substitute another valid expression (which can be the same as the first) for b in every place it occurs, then you will still end up with a true statement:
Substitute 2 for a and 3 for b (this is usually stated "let a = 2 and b = 3"):
23 + 33 = (2 + 3)(22 - (2)(3) + 32)
Let a = 2x and b = 3:
(2x)3 + 33 = (2x + 3)((2x)2 - (2x)(3) + 32)
let a = x+y and b = x-y:
(x+y)3 + (x-y)3 = ((x+y) + (x-y))((x+y)2 - (x+y)(x-y) + (x-y)2)
All of these can be checked individually to see if they are true, but by doing the algebra on the parametrized formula, a3 + b3 = (a + b)(a2 - ab + b2), we can prove that it is true regardless of what substitutions we use for a and b. This is why we use parameters.
Another example is the quadratic formula: if ax2 + bx + c = 0, then x = (-b +/- sqrt(b2 - 4ac))/2a.
In this relation, x is a variable; a, b, and c are parameters, and the point is to find the values of the variable x for which the expression ax2 + bx + c = 0.
For example, if we want to find the values for x that make x2 - 5x + 6 = 0, then we assign values to the parameters as follows:
first we rewrite the expression to have the same form as the one in the quadratic formula:
x2 - 5x + 6 = 1x2 + (-5)x + 6.
Then we assign values to the parameters a, b, c so that ax2 + bx + c = 1x2 + (-5)x + 6. We see that if we let
a = 1
b = -5
c = 6,
the two expressions will be equal.
Now we plug these values into the quadratic formula:
x = ( -(-5) +/- sqrt((-5)2 - 4(1)(6)) )/2(1)
= (5 +/- sqrt(25 - 24))/2
= (5 +/- 1)/2
x = (5+1)/2 = 3 or x = (5-1)/2 = 2
so, 2 and 3 are the two values that can be substituted for x in (x2 - 5x + 6) to make the overall value of the expression 0:
(2)2 - 5(2) + 6 = 4 - 10 + 6 = 0.
(3)2 - 5(3) + 6 = 9 - 15 + 6 = 0.
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Icarus
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Re: Factor Completely
« Reply #11 on: Apr 24th, 2006, 5:33pm » |
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As for synthetic division, it is a process that you clearly have not learned yet (if you had, you would have known immediately what I was refering to). Synthetic division is a fast way of dividing any polynomial by a linear divisor like 2x + 3 ("linear" means that it has only a constant term and a term in x - no x2 term or terms with higher powers).
You will learn about synthetic division later. If not in this course, then in your next algebra course.
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"Pi goes on and on and on ...
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I wonder: Which is larger
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Mugwump101
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Re: Factor Completely
« Reply #12 on: Apr 27th, 2006, 1:31am » |
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No one in my class got an answer for this extra credit question but my teacher taught us how to do it and the formula.
Also, I was reading ahead in textbook and an online website. I think I finally get what you're talking and thanks Icarus for explaining it, and that there are two ways to it or even more out there. I understand the difference bet. the quadratic and the sum/difference of two cubes, and complex numbers. I think Synthetic division I'll learn in the next course because I didn't find any material in my textbook on it.
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Icarus
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Re: Factor Completely
« Reply #13 on: Apr 27th, 2006, 2:46pm » |
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on Apr 27th, 2006, 1:31am, Mugwump101 wrote:| I understand the difference bet. the quadratic and the sum/difference of two cubes, and complex numbers. |
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I always prefer to over-explain than to under-explain. If I over-explain, then you just have to skim over it until I get to something you don't know. But if I under-explain, then you can't figure out my meaning, so my whole message is worthless.
Since I can't be sure what you know and don't know, I aim low.
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